Prove $$\lim_{x \rightarrow 3}x^2=9$$ via delta epsilon

This is what I have so far

$|x^2-9|<\epsilon$ and $0<|x-3|<\delta$

Let $z=x-3$.


$\delta(\delta+6)<\epsilon$ and we let $\delta \leq 1$

so then we have $\delta<\frac{\epsilon}{7}$

Then when I try to do the actual proof I have the following-

$|7x-21|<\epsilon$ which gets me nowhere.

  • 4
    $\begingroup$ Tip: $|7x-21| = 7|x-3|$. $\endgroup$ Oct 14, 2018 at 1:04

1 Answer 1


Let $\epsilon>0$. Then let $\delta=\min\{1,\frac{\epsilon}{7}\}$ . Then from $-1<x-3<1$ it follows that $5\leq x+3\leq 7$ so $|x+3|\leq 7$. Then $|x-3|<\delta$ and we have \begin{equation}|x^2-9|=|(x-3)(x+3)|\leq |x-3||x+3|<\frac{\epsilon}{7}7=\epsilon \end{equation}


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